Mathematics: A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 185 milligrams with s = 17.6 milligrams. A confidence interval of 173.8 mg μ 196.2 mg is constructed for the true mean cholesterol content of all such eggs. It was assumed that the population has a normal distribution. What confidence level does this interval represent?

It would be the figure that’s 6yd by 18yd because they use the same scale factor (2)...

A laboratory tested twelve chicken eggs and found

rihanna79

11.06.2021

The appropriate solution is "95%".

Step-by-step explanation:

The given values are:

Lower bound,

= 173.8

Upper bound,

= 196.2

Mean of cholesterol,

= 138 milligrams

S = 17.6 mg

Now,

The margin of error (E) will be:

= $\frac{Upper \ bound-Lower \ bound}{2}$

= $\frac{196.2-173.8}{2}$

= $\frac{22.4}{2}$

= 11.2

As we know,

⇒ $E=t_{\alpha 12}\times \frac{S}{\sqrt{n} }$

On putting the values, we get

⇒ $11.2=t_{\alpha 12}\times \frac{17.6}{\sqrt{12} }$

On applying cross-multiplication, we get

⇒ $t_{\alpha 12}=2.2044$

∴ $\frac{\alpha}{2} =0.025$

$\alpha = 0.025\times 2$

=0.05

hence,

The confidence level will be:

⇒ $CI=1-\alpha$

⇒ =1-0.05

⇒ =0.95

⇒ =95 %

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